Question

an honest die is rolled twice. give your answers to the following questions in fractions. (a) the probability of getting a 4 on both rolls is . (b) the probability of showing a 3 on the first roll and an even number on the second roll is . (hint: it may help to create a table similar to the one in question 4 of section 3.1, but showing pairs instead of sums.)

Explanation:

Step1: Determine total number of outcomes

When a die is rolled once, there are 6 possible outcomes. When rolled twice, by the multiplication - principle, the total number of outcomes is $6\times6 = 36$.

Step2: Calculate probability for part (a)

The event of getting a 4 on the first roll and a 4 on the second roll is a single outcome. So the probability $P(A)=\frac{1}{36}$.

Step3: Calculate probability for part (b)

The probability of getting a 3 on the first roll is $\frac{1}{6}$. The probability of getting an even number (2, 4, or 6) on the second roll is $\frac{3}{6}=\frac{1}{2}$. Since the two rolls are independent events, the probability of both events occurring is the product of their probabilities. So $P(B)=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}$.

Answer:

(a) $\frac{1}{36}$
(b) $\frac{1}{12}$